Question
Compute the limit $$\lim\limits_{(x,y) \to (0,0)} \frac {(x+y)^2}{3 \lvert{x}\lvert + \lvert {y} \lvert}.$$
Edited Working
My original answer revolved around considering different paths and I was not really getting anywhere. I have since been given some hints to use the squeeze theorem instead and have come up with the following:
$$\frac{x+y}{3} = \frac{(x+y)^2}{3(x+y)} \leq \frac{(x +y)^2}{3 \lvert x \lvert + \lvert y \lvert} \leq \frac{(\lvert x \lvert + \lvert y \lvert )^2}{\lvert x \lvert + \lvert y \lvert} = \lvert x \lvert + \lvert y \lvert$$
Now, by the squeeze theorem, since $$\lim\limits_{(x,y) \to (0,0)} \lvert x \lvert + \lvert y \lvert = 0$$ and $$\lim\limits_{(x,y) \to (0,0)} \frac {x + y }{3} = 0,$$ so $$\lim\limits_{(x,y) \to (0,0)} \frac {(x+y)^2}{3 \lvert{x}\lvert + \lvert {y} \lvert} = 0.$$
As I have only just covered multi-variable calculus, I am wondering whether my approach and answer is correct.
HINT: $$ \frac{(x+y)^2}{3|x| + |y|} \le \frac{(x + y)^2}{|x| + |y|} \le \frac{x^2}{|x| + |y|} + \frac{2|xy|}{|x| + |y|} + \frac{y^2}{|x| + |y|} \le $$ $$ \le \frac{x^2}{|x|} + 2\frac{|xy|}{|y|} + \frac{y^2}{|y|}. $$ Of course, cases where $x$ or $y$ are equal to $0$ can be handled separately.